An oxygen cylinder of volume 30 litre has 18.20 moles of oxygen. After some oxygen is withdrawn from

Question

An oxygen cylinder of volume 30 litre has 18.20 moles of oxygen. After some oxygen is withdrawn from the cylinder, its gauge pressure drops to 11 atmospheric pressure at temperature $27^{\circ} \mathrm{C}$. The mass of the oxygen withdrawn from the cylinder is nearly equal to [Given, $R=\frac{100}{12} \mathrm{~J} \mathrm{~mol}^{-1} \mathrm{~K}^{-1}$, and molecular mass of $\mathrm{O}_2=32,1 \mathrm{~atm}$ pressure $=1.01 	imes 10^5 \mathrm{~N} / \mathrm{m}$ ]

Professor PC Thomas Classes · Answer

(a) : Initially volume of the oxygen cylinder,\ $ V_1=30 \mathrm{~L}=30 	imes 10^{-3} \mathrm{~m}^3 $ \Number of moles, $n_1=18.20$ $ T_1=27+273=300 \mathrm{~K} \ $ Final gauge pressure, $P_g=11 \mathrm{~atm}$ \ Final absolute pressure, $P_2=P_g+P_{\mathrm{atm}}=11 \mathrm{~atm}+1 \mathrm{~atm}$ $ =12 \mathrm{~atm} \ $ For oxygen, molecular weight, $M=32 \mathrm{~g}$ \ Initial mass of oxygen in the cylinder, $ m_1=n_1 M=18.20 	imes 32=582.4 \mathrm{~g} $ \Now, $V_2=30 	imes 10^{-3} \mathrm{~m}^3, P_2=12 	imes 1.01 	imes 10^5 \mathrm{~Pa}$ $ n_2=\frac{P_2 V_2}{R T_2}=\frac{\left(12 	imes 1.01 	imes 10^5ight) 	imes\left(30 	imes 10^{-3}ight)}{8.3 	imes 300}=14.60 $ \ ∴ $\quad$ Final mass of oxygen gas in the cylinder, \ $ m_2=14.60 	imes 32=467.2 \mathrm{~g} $ $	herefore \quad$ \ Mass of the withdrawn oxygen gas \ $ \begin{aligned} & =m_1-m_2 \ & =582.4-467.2 \ & =115.2 \mathrm{~kg} \approx 0.116 \mathrm{~kg} \end{aligned} $